If 0.44 g of substance dissolved in 22.2 g of benzene lowers the freezing point of benzene by 0.567°C, then the molecular mass of a substance is, (the molal depression constant = 5.12°C mol–1 kg)
Answer & explanation
Correct answer: option 3
The correct answer is option 3. 178.9.
Given,
Weight of substance \(= 0.440 g\)
Weight of substance \(= 22.2 g\)
\(\Delta T_f = 0.567^o\text{C kg mol}^{-1}\)
\(K_f = 5.12^oC\)
We know,
Depression in freezing point,
\(\Delta T_f = K_f × m\)
\(⇒ 0.567^oC = 5.12^o\text{C kg mol}^{-1} × \frac{0.44 × 1000}{M × 22.2 g}\)
\(⇒ M = \frac{5.12 × 0.44 ×1000}{22.2 × 0.567}\)
\(⇒ M = \frac{2252.8}{12.58}\)
or, \(M = 178.9\)